Solve (1+i)(2+2i)/(2-2i)(2+2i)=

$$\frac { ( 1 + i ) ( 2 + 2 i ) } { ( 2 - 2 i ) ( 2 + 2 i ) } =$$

Evaluate

$$\frac{1}{2}i=0.5i$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(1+i\right)\left(2+2i\right)}{2^{2}-2^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{\left(1+i\right)\left(2+2i\right)}{8}$$

Multiply complex numbers 1+i and 2+2i like you multiply binomials.

$$\frac{1\times 2+1\times \left(2i\right)+2i+2i^{2}}{8}$$

By definition, i^{2} is -1.

$$\frac{1\times 2+1\times \left(2i\right)+2i+2\left(-1\right)}{8}$$

Do the multiplications in 1\times 2+1\times \left(2i\right)+2i+2\left(-1\right).

$$\frac{2+2i+2i-2}{8}$$

Combine the real and imaginary parts in 2+2i+2i-2.

$$\frac{2-2+\left(2+2\right)i}{8}$$

Do the additions in 2-2+\left(2+2\right)i.

$$\frac{4i}{8}$$

Divide 4i by 8 to get \frac{1}{2}i.

$$\frac{1}{2}i$$

Real Part

$$0$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{\left(1+i\right)\left(2+2i\right)}{2^{2}-2^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{\left(1+i\right)\left(2+2i\right)}{8})$$

Multiply complex numbers 1+i and 2+2i like you multiply binomials.

$$Re(\frac{1\times 2+1\times \left(2i\right)+2i+2i^{2}}{8})$$

By definition, i^{2} is -1.

$$Re(\frac{1\times 2+1\times \left(2i\right)+2i+2\left(-1\right)}{8})$$

Do the multiplications in 1\times 2+1\times \left(2i\right)+2i+2\left(-1\right).

$$Re(\frac{2+2i+2i-2}{8})$$

Combine the real and imaginary parts in 2+2i+2i-2.

$$Re(\frac{2-2+\left(2+2\right)i}{8})$$

Do the additions in 2-2+\left(2+2\right)i.

$$Re(\frac{4i}{8})$$

Divide 4i by 8 to get \frac{1}{2}i.

$$Re(\frac{1}{2}i)$$

The real part of \frac{1}{2}i is 0.

$$0$$